question_answer

Six long parallel current carrying wires are perpendicular to the plane of the figure. They pass through the vertices of a regular hexagon of side length a. All wires have same current /. Direction of current is out of the plane of the figure in all the wires except the one passing through vertex F, which has current directed into the plane of the figure. The magnitude and direction of magnetic induction field at the centre of the hexagon will be

A) \[\frac{{{\mu }_{0}}I}{\pi a}\]towards midpoint of side DE   

B)  \[\frac{3{{\mu }_{0}}I}{\pi a}\] towards midpoint of side BC

C) \[\frac{3{{\mu }_{0}}I}{\pi a}\]towards midpoint of side DE                 

D) \[\frac{{{\mu }_{0}}I}{\pi a}\] towards midpoint of side BC

Correct Answer: A

Solution :

[a] Imagine a current I at F in the same direction as that of other 5 currents. Such 6 currents will produce zero field at the centre. Hence field at the centre will be due to a current \[2I\] at F directed into the plane of the figure.             \[B=\frac{{{\mu }_{0}}(2I)}{2\pi a}=\frac{{{\mu }_{0}}I}{\pi a}\] The direction is perpendicular to the line\[FO\]. This is a direction towards midpoint of side DE.